math t 30/3/2012题目更新

puangenlun

integral sin^4(2 x) dx
For the integrand sin^4(2 x), substitute u = 2 x and  du = 2 dx:
= 1/2 integral sin^4(u) du
Use the reduction formula,  integral sin^m(u) du = -(cos(u) sin^(m-1)(u))/m + (m-1)/m integral sin^(-2+m)(u) du, where m = 4:
= 3/8 integral sin^2(u) du-1/8 sin^3(u) cos(u)
Write sin^2(u) as 1/2-1/2 cos(2 u):
= 3/8 integral (1/2-1/2 cos(2 u)) du-1/8 sin^3(u) cos(u)
Integrate the sum term by term and factor out constants:
= -1/8 sin^3(u) cos(u)+3/8 integral 1/2 du-3/16 integral cos(2 u) du
For the integrand cos(2 u), substitute s = 2 u and  ds = 2 du:
= -3/32 integral cos(s) ds-1/8 sin^3(u) cos(u)+3/8 integral 1/2 du
The integral of 1/2 is u/2:
= -3/32 integral cos(s) ds+(3 u)/16-1/8 sin^3(u) cos(u)
The integral of cos(s) is sin(s):
= -(3 sin(s))/32+(3 u)/16-1/8 sin^3(u) cos(u)+constant
Substitute back for s = 2 u:
= (3 u)/16-3/32 sin(2 u)-1/8 sin^3(u) cos(u)+constant
Substitute back for u = 2 x:
= (3 x)/8-3/32 sin(4 x)-1/8 sin^3(2 x) cos(2 x)+constant
Which is equal to:
= 1/64 (24 x-8 sin(4 x)+sin(8 x))+constant

whyyie

回复 20# jypang

sin^2 2x = 1 - cos^2 2x
(sin^2 2x)^2 = (1-cos^2 2x)(sin^2 2x) = sin^2 2x  - sin^2 2x cos^2 2x = sin^2 2x - (sin 2x cos 2x)^2

To change  sin^2 2x
From cos 2x = 1 - 2 sin^2 x
2 sin^2 x = 1 - cos 2x
2 sin^2 2x = 1 - cos 4x
sin^2 2x = (1/2)(1-cos 4x)

To change (sin 2x cos 2x)^2
From sin 2x = 2 sin x cos x
sin 4x = 2 sin 2x cos 2x
(1/2) sin 4x = sin 2x cos 2x
(sin 2x cos 2x)^2 = (1/4) sin^2 4x

From 2 sin^2 x = 1 - cos 2x
sin^2 4x  = (1/2)(1-cos 8x)
(1/4)(sin^2 4x) = (1/8) ( 1- cos 8x)

Integration of (sin^2 2x)^2
= Integration of  sin^2 2x -  Integration of (sin 2x cos 2x)^2
= Integration of (1/2)(1-cos 4x) - Integration of (1/8) ( 1- cos 8x)
= (1/64) (24x-8 sin 4x +sin 8x)+c

jypang

integral sin^4(2 x) dx
For the integrand sin^4(2 x), substitute u = 2 x and  du = 2 dx:
= 1/2 int ...
puangenlun 发表于 4-2-2012 09:52 PM

回复  jypang

sin^2 2x = 1 - cos^2 2x
(sin^2 2x)^2 = (1-cos^2 2x)(sin^2 2x) = sin^2 2x  - sin^2  ...
whyyie 发表于 4-2-2012 11:45 PM

thanks a lot !!!!

jypang

Allmaths

jypang 发表于 30-3-2012 11:23 AM

    先找alpha 和Q的关系，得到

tan (alpha)=(AO/MO)tan Q           ,         (O 是square base 的中心, M 是 midpoint of line AB)

再prove AO:MO的比列是 (2)^(1/2) : 1

提示：pythagoras theorem

(a)同样的， 找beta和alpha的关系，得到

tan (beta)=(MO/AM) sec (alpha)

再prove MO:AM的比列是1 : 1

提示:AB=BC=CD=AD

(b)题目不完整

jypang

先找alpha 和Q的关系，得到

tan (alpha)=(AO/MO)tan Q           ,         (O 是square base ...
Allmaths 发表于 1-4-2012 03:41 PM

    题目里面的问题和a)我都做了，b)不会做，题目没错啊

Allmaths

题目里面的问题和a)我都做了，b)不会做，题目没错啊
jypang 发表于 1-4-2012 07:38 PM

    cos (gamma)= - cos2(alpha)?

还是，

cos (gamma)= - cos (2/?)(alpha)?

看回你的照片，好像有typo

jypang

cos (gamma)= - cos2(alpha)?

还是，

cos (gamma)= - cos (2/?)(alpha)?

看回你的照片， ...
Allmaths 发表于 1-4-2012 08:20 PM

    cos (gamma)= - cos ^2(alpha)
那个格子不小心弄到的